# Work pumping oil out of partially filled horizontal tank

1. Sep 15, 2011

### Phox

1. The problem statement, all variables and given/known data

"Oil weighing 9000N/m3 is to be pumped out of the top of a horizontal cylindrical tank of radius 5m and length 15m. Set up an integral to find the work required to pump out the oil to the top of the tank if the depth of the oil is 4m."

2. Relevant equations
I used y = (r2-x2)(1/2) for the equation of a semicircle.

3. The attempt at a solution
First I decided to try and find the area of one vertical "slice" of the oil. So what I've done is graphed a semicircle of radius 5 to represent this. The equation was x = (52-y2)(1/2). I used dy as my variable of integration. And then I said it was bounded from 1 to 5 (given that the depth of the oil is 4.)

From this exact integral I got the answer 19.56m2. I then multiplied this by 15m to get the volume - 293.49m3.

Given that the density of the oil is 9000N/m3, I found the weight of the oil to be 2,641,410N.

Now I need to set up the integral to find the work done. This is where I'm stuck.

2. Sep 15, 2011

### Phox

This might help to visualize it a bit

3. Sep 16, 2011

### danielakkerma

Hi,
Firstly, let me say that on the face of it, it seems you've done a thorough job, well done!
But you neglected one key factor, and that's the work itself?
Why is there work to be done anyway, it's because gravity resists the extraction of the oil, whic has weight.
Needless to say, $W = \displaystyle \int \vec{F} \cdot d\vec{r}$
Every ounce removed, for every inch, is work done.
So you have to incorporate that into your integral, as I see it.
Daniel